In a parallelogram ABCD, if vertices are in respective order and diagonals AC and BD intersect at O, then which of the following is NOT always correct?
- AΔAOB≡ΔCOD
- BΔAOD≡ΔCOD
- CΔABC≡ΔADC
- DΔBOC≡ΔAOD
Solution & Step-by-step Explanation
In any parallelogram ABCD where diagonals AC and BD intersect at O:
Diagonals bisect each other, so AO=CO and BO=DO.
Opposite sides are equal, so AB=CD and AD=BC.
Let's evaluate the congruence options using standard side-side-side (SSS) or side-angle-side (SAS) properties:
Option A: In ΔAOB and ΔCOD, AO=CO, BO=DO, and AB=CD. Thus, ΔAOB≡ΔCOD is always true.
Option C: In ΔABC and ΔADC, AB=CD, BC=DA, and AC=CA. Thus, ΔABC≡ΔADC is always true.
Option D: In ΔBOC and ΔAOD, BO=DO, CO=AO, and BC=AD. Thus, ΔBOC≡ΔAOD is always true.
Option B: In ΔAOD and ΔCOD, AO=CO and OD=OD, but AD is not necessarily equal to CD unless the parallelogram is a rhombus. Therefore, ΔAOD≡ΔCOD is NOT always correct.
Diagonals bisect each other, so AO=CO and BO=DO.
Opposite sides are equal, so AB=CD and AD=BC.
Let's evaluate the congruence options using standard side-side-side (SSS) or side-angle-side (SAS) properties:
Option A: In ΔAOB and ΔCOD, AO=CO, BO=DO, and AB=CD. Thus, ΔAOB≡ΔCOD is always true.
Option C: In ΔABC and ΔADC, AB=CD, BC=DA, and AC=CA. Thus, ΔABC≡ΔADC is always true.
Option D: In ΔBOC and ΔAOD, BO=DO, CO=AO, and BC=AD. Thus, ΔBOC≡ΔAOD is always true.
Option B: In ΔAOD and ΔCOD, AO=CO and OD=OD, but AD is not necessarily equal to CD unless the parallelogram is a rhombus. Therefore, ΔAOD≡ΔCOD is NOT always correct.